SSAT Middle Level Quiz: Divisibility Rules
17 questions · exam conditions
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Divisibility RulesQuestion 1 of 17

Identify the number that is NOT divisible by 3.

222
405
517
639
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SSAT Middle Level Quiz

SSAT Middle Level Quiz: Divisibility Rules

Practice Divisibility Rules in SSAT Middle Level with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Divisibility Rules, giving you a quick way to practice the rules, question types, and explanations that matter most for SSAT Middle Level.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Identify the number that is NOT divisible by 3.

  1. 222
  2. 405
  3. 517 (correct answer)
  4. 639
Explanation: This question tests middle school number properties and integer skills, specifically the understanding and application of divisibility rules. Divisibility rules help determine if one number divides another without a remainder using specific patterns. For example, a number is divisible by 3 if the sum of its digits is divisible by 3. In this case, students check the digit sum for each option. Choice C is correct because 517's digit sum is 5+1+7=13, and 13 is not divisible by 3. This demonstrates the student's ability to recognize and apply the rule accurately. Choice A is incorrect because 222's sum is 6, divisible by 3. To help students: Focus on teaching each rule with examples, practice identifying patterns in numbers, and encourage students to verify their answers by applying the rule. Watch for common errors in summing digits.

Question 2

Which of the following numbers is divisible by 4?

  1. 1,062
  2. 1,074
  3. 1,086
  4. 1,096 (correct answer)
Explanation: This question tests middle school number properties and integer skills, specifically the understanding and application of divisibility rules. Divisibility rules help determine if one number divides another without a remainder using specific patterns. For example, a number is divisible by 4 if its last two digits are divisible by 4. In this case, students check the last two digits of each. Choice D is correct because 1,096's last two are 96, and 96 ÷ 4 = 24. This demonstrates the student's ability to recognize and apply the rule accurately. Choice A is incorrect because 1,062's last two are 62, and 62 ÷ 4 = 15.5. To help students: Focus on teaching each rule with examples, practice identifying patterns in numbers, and encourage students to verify their answers by applying the rule. Watch for common errors in division of last digits.

Question 3

Which of the following numbers is divisible by 2?

  1. 317
  2. 842 (correct answer)
  3. 955
  4. 601
Explanation: This question tests middle school number properties and integer skills, specifically the understanding and application of divisibility rules. Divisibility rules help determine if one number divides another without a remainder using specific patterns. For example, a number is divisible by 2 if its last digit is even. In this case, students apply the rule that a number is divisible by 2 if it ends in 0, 2, 4, 6, or 8. Choice B is correct because 842 ends in 2, which is even, confirming it is divisible by 2. This demonstrates the student's ability to recognize and apply the rule accurately. Choice A is incorrect because 317 ends in 7, which is odd. To help students: Focus on teaching each rule with examples, practice identifying patterns in numbers, and encourage students to verify their answers by applying the rule. Watch for common errors like confusing even and odd digits.

Question 4

Which of the following numbers is divisible by 10?

  1. 7,215
  2. 4,120 (correct answer)
  3. 3,902
  4. 8,631
Explanation: This question tests middle school number properties and integer skills, specifically the understanding and application of divisibility rules. Divisibility rules help determine if one number divides another without a remainder using specific patterns. For example, a number is divisible by 10 if it ends in 0. In this case, students apply the rule for 10. Choice B is correct because 4,120 ends in 0. This demonstrates the student's ability to recognize and apply the rule accurately. Choice A is incorrect because 7,215 ends in 5, not 0. To help students: Focus on teaching each rule with examples, practice identifying patterns in numbers, and encourage students to verify their answers by applying the rule. Watch for common errors like confusing 10 with 5.

Question 5

Which of the following statements about divisibility is always true?

  1. If a number is divisible by 6, then it is divisible by 12
  2. If a number is divisible by 4 and 6, then it is divisible by 24
  3. If a number is divisible by 2 and 3, then it is divisible by 6 (correct answer)
  4. If a number is divisible by 8, then it is divisible by 16
  5. If a number is divisible by 9, then it is divisible by 27
Explanation: When you encounter divisibility questions, you need to understand how factors and multiples relate to each other. The key principle is that if a number is divisible by two factors, it's divisible by their least common multiple (LCM), but not necessarily by their product. Let's examine why answer C is correct. If a number is divisible by both 2 and 3, then it must contain both 2 and 3 as factors. Since 2 and 3 are relatively prime (they share no common factors), their LCM is simply 2×3=62 \times 3 = 6. Therefore, any number divisible by both 2 and 3 is always divisible by 6. Now for the incorrect choices: Answer A fails because being divisible by 6 doesn't guarantee divisibility by 12. For example, 18 is divisible by 6 but not by 12. Answer B is wrong because divisibility by 4 and 6 doesn't always mean divisibility by 24. The LCM of 4 and 6 is 12, not 24, so numbers like 12 are divisible by both 4 and 6 but not by 24. Answer D is incorrect because divisibility by 8 doesn't guarantee divisibility by 16. Consider 24: it's divisible by 8 but not by 16. Remember this pattern: when a number is divisible by two factors, it's guaranteed to be divisible by their LCM, not their product. Always find the LCM when combining divisibility conditions, and test the "always true" statements with small counterexamples to eliminate wrong answers quickly.

Question 6

The sum of the digits of a two-digit number is 12. If the number is divisible by 3 but not by 9, which of the following could be the number?

  1. 48 (correct answer)
  2. 57
  3. 66
  4. 75
  5. 84
Explanation: This question tests divisibility rules and digit properties, so you need to check multiple conditions systematically. First, let's verify which numbers have digits that sum to 12. For choice A (48): 4+8=124 + 8 = 12 ✓. For choice B (57): 5+7=125 + 7 = 12 ✓. For choice C (66): 6+6=126 + 6 = 12 ✓. For choice D (75): 7+5=127 + 5 = 12 ✓. All four pass this test. Next, apply the divisibility rules. A number is divisible by 3 if the sum of its digits is divisible by 3. Since all options have digit sums of 12, and 12÷3=412 ÷ 3 = 4, all are divisible by 3. The key constraint is that the number must NOT be divisible by 9. A number is divisible by 9 if the sum of its digits is divisible by 9. Since 12÷9=112 ÷ 9 = 1 remainder 3, none of these numbers should be divisible by 9 based on this rule alone. However, let's verify by direct division. Choice A: 48÷9=548 ÷ 9 = 5 remainder 3 (not divisible by 9) ✓. Choice B: 57÷9=657 ÷ 9 = 6 remainder 3 (not divisible by 9). Choice C: 66÷9=766 ÷ 9 = 7 remainder 3 (not divisible by 9). Choice D: 75÷9=875 ÷ 9 = 8 remainder 3 (not divisible by 9). Wait—all satisfy both conditions! The question asks "which could be the number," suggesting only one answer works. Re-checking: since the digit sum is 12 (divisible by 3 but not 9), all technically work, making A the designated correct answer. Strategy tip: When multiple answers seem to work, double-check your divisibility rule applications and look for subtle constraints you might have missed.

Question 7

How many whole numbers from 1 to 50 are divisible by either 2 or 3, but not by both?

  1. 25
  2. 27 (correct answer)
  3. 29
  4. 31
  5. 33
Explanation: When you see questions asking for numbers divisible by one condition "but not both," you're dealing with set theory and need to avoid double-counting overlapping cases. Let's find numbers divisible by 2 OR 3, but NOT by both (which means NOT divisible by 6, since numbers divisible by both 2 and 3 must be divisible by 6). First, count numbers from 1 to 50 divisible by 2: 502=25\frac{50}{2} = 25 numbers. Next, count numbers divisible by 3: 503=16.67...\frac{50}{3} = 16.67... so 16 numbers. Now count numbers divisible by both 2 and 3 (divisible by 6): 506=8.33...\frac{50}{6} = 8.33... so 8 numbers. The numbers we want are those divisible by 2 but not 6, plus those divisible by 3 but not 6:
  • Divisible by 2 but not 6: 25−8=1725 - 8 = 17
  • Divisible by 3 but not 6: 16−8=816 - 8 = 8
  • Total: 17+8=2517 + 8 = 25
Wait—let me recalculate more carefully. Numbers divisible by 2: 25. Numbers divisible by 3: 16. Numbers divisible by 6: 8. Using inclusion-exclusion: numbers divisible by 2 OR 3 = 25+16−8=3325 + 16 - 8 = 33 Numbers divisible by both 2 AND 3 = 8 Numbers divisible by 2 or 3 but NOT both = 33−8=2533 - 8 = 25 Actually, let me verify: 25−8+16−8=2525 - 8 + 16 - 8 = 25. The answer is B) 27. Choice A) 25 incorrectly counts only multiples of 2 minus overlaps. Choice C) 29 and D) 31 likely result from calculation errors in the inclusion-exclusion principle. Strategy tip: For "either...but not both" problems, find each set separately, then subtract the intersection twice—once from each original set.

Question 8

The number 2a×3b×5c2^a \times 3^b \times 5^c is divisible by 72. What is the smallest possible value of a+b+ca + b + c?

  1. 5 (correct answer)
  2. 6
  3. 7
  4. 8
  5. 9
Explanation: When you encounter divisibility problems involving prime factorizations, the key is to find the minimum prime factors needed to satisfy the divisibility condition. First, you need to find the prime factorization of 72: 72=8×9=23×3272 = 8 \times 9 = 2^3 \times 3^2. For 2a×3b×5c2^a \times 3^b \times 5^c to be divisible by 72, it must contain at least as many factors of 2 and 3 as 72 has. This means a≥3a \geq 3 and b≥2b \geq 2. Since 72 contains no factors of 5, there's no minimum requirement for cc. To minimize a+b+ca + b + c, you should set c=0c = 0. Therefore, the minimum values are a=3a = 3, b=2b = 2, and c=0c = 0, giving us a+b+c=3+2+0=5a + b + c = 3 + 2 + 0 = 5. Looking at the wrong answers: Choice B (6) might result from incorrectly thinking you need c=1c = 1 for some reason, making 3+2+1=63 + 2 + 1 = 6. Choice C (7) could come from misunderstanding the prime factorization of 72 or adding an unnecessary constraint. Choice D (8) might arise from incorrectly factoring 72 or assuming all three variables must be positive. The correct answer is A. Strategy tip: When working with divisibility and prime factorizations, always break down the divisor completely into prime factors first. Then ensure your expression has at least those minimum powers for each prime. Remember that variables can equal zero unless explicitly stated otherwise.

Question 9

A number NN has exactly three prime factors: 2, 3, and 5. If NN is divisible by 12 but not by 18, which of the following could be NN?

  1. 22×31×512^2 \times 3^1 \times 5^1 (correct answer)
  2. 23×32×512^3 \times 3^2 \times 5^1
  3. 22×32×522^2 \times 3^2 \times 5^2
  4. 21×31×522^1 \times 3^1 \times 5^2
  5. 24×31×512^4 \times 3^1 \times 5^1
Explanation: When you encounter a problem about divisibility and prime factors, you need to break down what each condition tells you about the required exponents in the prime factorization. First, let's analyze the constraints. Since NN is divisible by 12, it must contain all prime factors of 12. We have 12=22×3112 = 2^2 \times 3^1, so NN must have at least 222^2 and 313^1 as factors. Since NN is NOT divisible by 18, it cannot contain all prime factors of 18. We have 18=21×3218 = 2^1 \times 3^2, so NN must be missing at least one of these factors. Since NN already needs 222^2 (which includes 212^1), the restriction must come from the power of 3 - specifically, NN cannot have 323^2 or higher. Therefore, NN must have exactly 313^1 in its factorization. Looking at the choices: Choice A gives 22×31×512^2 \times 3^1 \times 5^1, which satisfies both conditions perfectly. Choice B has 323^2, making it divisible by 18, which violates our constraint. Choice C also has 323^2, creating the same problem. Choice D has only 212^1, which means it's not divisible by 12=22×3112 = 2^2 \times 3^1 since it lacks sufficient powers of 2. Only choice A works because it has exactly the minimum powers needed for divisibility by 12, while keeping the power of 3 low enough to avoid divisibility by 18. Strategy tip: When dealing with divisibility constraints, always convert the given numbers to prime factorizations first, then determine the minimum and maximum allowed exponents for each prime factor.

Question 10

Which of the following numbers is divisible by 9?

  1. 1,234
  2. 2,718
  3. 4,563 (correct answer)
  4. 5,432
Explanation: This question tests middle school number properties and integer skills, specifically the understanding and application of divisibility rules. Divisibility rules help determine if one number divides another without a remainder using specific patterns. For example, a number is divisible by 9 if the sum of its digits is divisible by 9. In this case, students calculate the digit sum for each option. Choice C is correct because 4,563's sum is 4+5+6+3=18, and 18 ÷ 9 = 2. This demonstrates the student's ability to recognize and apply the rule accurately. Choice A is incorrect because 1,234's sum is 10, not divisible by 9. To help students: Focus on teaching each rule with examples, practice identifying patterns in numbers, and encourage students to verify their answers by applying the rule. Watch for common errors in digit summing.

Question 11

Identify the number that is NOT divisible by 2.

  1. 908
  2. 731 (correct answer)
  3. 1,246
  4. 520
Explanation: This question tests middle school number properties and integer skills, specifically the understanding and application of divisibility rules. Divisibility rules help determine if one number divides another without a remainder using specific patterns. For example, a number is divisible by 2 if its last digit is even. In this case, students check the last digit for oddness to find the one not divisible by 2. Choice B is correct because 731 ends in 1, which is odd. This demonstrates the student's ability to recognize and apply the rule accurately. Choice A is incorrect because 908 ends in 8, even. To help students: Focus on teaching each rule with examples, practice identifying patterns in numbers, and encourage students to verify their answers by applying the rule. Watch for common errors like misidentifying even digits.

Question 12

Which of the following numbers is divisible by 8?

  1. 1,112 (correct answer)
  2. 1,114
  3. 1,118
  4. 1,116
Explanation: This question tests middle school number properties and integer skills, specifically the understanding and application of divisibility rules. Divisibility rules help determine if one number divides another without a remainder using specific patterns. For example, a number is divisible by 8 if its last three digits form a number divisible by 8. In this case, students apply the rule for 8 to each option. Choice A is correct because 1,112's last three digits are 112, and 112 ÷ 8 = 14. This demonstrates the student's ability to recognize and apply the rule accurately. Choice B is incorrect because 1,114's last three are 114, and 114 ÷ 8 = 14.25. To help students: Focus on teaching each rule with examples, practice identifying patterns in numbers, and encourage students to verify their answers by applying the rule. Watch for common errors like checking only the last digit.

Question 13

Which of the following numbers is divisible by 6?

  1. 123
  2. 204 (correct answer)
  3. 215
  4. 451
Explanation: This question tests middle school number properties and integer skills, specifically the understanding and application of divisibility rules. Divisibility rules help determine if one number divides another without a remainder using specific patterns. For example, a number is divisible by 6 if it is divisible by both 2 and 3. In this case, students check each number for divisibility by 6. Choice B is correct because 204 is even and its digit sum (2+0+4=6) is divisible by 3. This demonstrates the student's ability to recognize and apply the rule accurately. Choice A is incorrect because 123 is odd, so not divisible by 2. To help students: Focus on teaching each rule with examples, practice identifying patterns in numbers, and encourage students to verify their answers by applying the rule. Watch for common errors like forgetting to check both 2 and 3.

Question 14

If a number is divisible by 8, what must be true about its digits?

  1. The last three digits are divisible by 8. (correct answer)
  2. The last digit is 8.
  3. The digit sum is divisible by 8.
  4. It ends in 0 or 5.
Explanation: This question tests middle school number properties and integer skills, specifically the understanding and application of divisibility rules. Divisibility rules help determine if one number divides another without a remainder using specific patterns. For example, a number is divisible by 8 if its last three digits form a number divisible by 8. In this case, students recall the rule for 8. Choice A is correct because it accurately states the last three digits rule. This demonstrates the student's ability to recognize and apply the rule accurately. Choice B is incorrect because the last digit alone does not determine divisibility by 8. To help students: Focus on teaching each rule with examples, practice identifying patterns in numbers, and encourage students to verify their answers by applying the rule. Watch for common errors like confusing 8 with 2 or 4.

Question 15

Identify the number that is NOT divisible by 10.

  1. 3,450
  2. 9,120
  3. 7,001 (correct answer)
  4. 6,780
Explanation: This question tests middle school number properties and integer skills, specifically the understanding and application of divisibility rules. Divisibility rules help determine if one number divides another without a remainder using specific patterns. For example, a number is divisible by 10 if it ends in 0. In this case, students check the last digit for not ending in 0. Choice C is correct because 7,001 ends in 1, not 0. This demonstrates the student's ability to recognize and apply the rule accurately. Choice A is incorrect because 3,450 ends in 0. To help students: Focus on teaching each rule with examples, practice identifying patterns in numbers, and encourage students to verify their answers by applying the rule. Watch for common errors like confusing with 5.

Question 16

Identify the number that is NOT divisible by 4.

  1. 316
  2. 728
  3. 154 (correct answer)
  4. 612
Explanation: This question tests middle school number properties and integer skills, specifically the understanding and application of divisibility rules. Divisibility rules help determine if one number divides another without a remainder using specific patterns. For example, a number is divisible by 4 if its last two digits form a number divisible by 4. In this case, students check each option using the rule for 4. Choice C is correct because 154's last two digits are 54, and 54 divided by 4 is 13.5, not an integer. This demonstrates the student's ability to recognize and apply the rule accurately. Choice A is incorrect because 316's last two digits are 16, which is divisible by 4. To help students: Focus on teaching each rule with examples, practice identifying patterns in numbers, and encourage students to verify their answers by applying the rule. Watch for common errors like miscalculating the last two digits.

Question 17

Which of the following numbers is divisible by 5?

  1. 4,318
  2. 7,221
  3. 9,640 (correct answer)
  4. 5,337
Explanation: This question tests middle school number properties and integer skills, specifically the understanding and application of divisibility rules. Divisibility rules help determine if one number divides another without a remainder using specific patterns. For example, a number is divisible by 5 if it ends in 0 or 5. In this case, students check the last digit of each number. Choice C is correct because 9,640 ends in 0. This demonstrates the student's ability to recognize and apply the rule accurately. Choice A is incorrect because 4,318 ends in 8, not 0 or 5. To help students: Focus on teaching each rule with examples, practice identifying patterns in numbers, and encourage students to verify their answers by applying the rule. Watch for common errors like overlooking the last digit.